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Option.fold() considered unreadable

We had a lengthy discussion recently during code review whether scala.Option.fold() is idiomatic and clever or maybe unreadable and tricky? Let's first describe what the problem is. Option.fold does two things: maps a function f over Option's value (if any) or returns an alternative alt if it's absent. Using simple pattern matching we can implement it as follows:

If you prefer one-liner, fold is actually a combination of map and getOrElse

val x: R = option map f getOrElse alt

Or, if you are a C programmer that still wants to write in C, but using Scala compiler:

val x: R = if (option.isDefined)
f(option.get)
else
alt

Interestingly this is similar to how fold() is actually implemented, but that's an implementation detail. OK, all of the above can be replaced with single Option.fold():

val x: R = option.fold(alt)(f)

Technically you can even use /: and \: operators (alt /: option) - but that would be simply masochistic. I have three problems with option.fold() idiom. First of all - it's anything but readable. We are folding (reducing) over Option - which doesn't really make much sense. Secondly it reverses the ordinary positive-then-negative-case flow by starting with failure (absence, alt) condition followed by presence block (f function; see also: Refactoring map-getOrElse to fold). Interestingly this method would work great for me if it was named mapOrElse:

"I personally find methods like cata that take two closures as arguments are often overdoing it. Do you really gain in readability over map + getOrElse? Think of a newcomer to your code[...]"

While cata has some theoretical background, Option.fold just sounds like a random name collision that doesn't bring anything to the table, apart from confusion. I know what you'll say, that TraversableOnce has fold and we are sort-of doing the same thing. Why it's a random collision rather than extending the contract described in TraversableOnce? fold() method in Scala collections typically just delegates to one of foldLeft()/foldRight() (the one that works better for given data structure), thus it doesn't guarantee order and folding function has to be associative. But in Option.fold() the contract is different: folding function takes just one parameter rather than two. If you read my previous article about folds you know that reducing function always takes two parameters: current element and accumulated value (initial value during first iteration). But Option.fold() takes just one parameter: current Option value! This breaks the consistency, especially when realizing Option.foldLeft() and Option.foldRight() have correct contract (but it doesn't mean they are more readable).

The only way to understand folding over option is to imagine Option as a sequence with 0 or 1 elements. Then it sort of makes sense, right? No.

If we treat Option[T] as a List[T], awkward Option.fold() breaks because it has different type than TraversableOnce.fold(). This is my biggest concern. I can't understand why folding wasn't defined in terms of the type system (trait?) and implemented strictly. As an example take a look at:

Data.Foldable in Haskell (advanced)

Data.Foldable typeclass describes various flavours of folding in Haskell. There are familiar foldl/foldr/foldl1/foldr1, in Scala named foldLeft/foldRight/reduceLeft/reduceRight accordingly. They have the same type as Scala and behave unsurprisingly with all types that you can fold over, including Maybe, lists, arrays, etc. There is also a function named fold, but it has a completely different meaning:

class Foldable t where
fold :: Monoid m => t m -> m

While other folds are quite complex, this one barely takes a foldable container of ms (which have to be Monoids) and returns the same Monoid type. A quick recap: a type can be a Monoid if there exists a neutral value of that type and an operation that takes two values and produces just one. Applying that function with one of the arguments being neutral value yields the other argument. String ([Char]) is a good example with empty string being neutral value (mempty) and string concatenation being such operation (mappend). Notice that there are two different ways you can construct monoids for numbers: under addition with neutral value being 0 (x + 0 == 0 + x == x for any x) and under multiplication with neutral 1 (x * 1 == 1 * x == x for any x). Let's stick to strings. If I fold empty list of strings, I'll get an empty string. But when a list contains many elements, they are being concatenated:

In the first example we have to explicitly say what is the type of empty list []. Otherwise Haskell compiler can't figure out what is the type of elements in a list, thus which monoid instance to choose. In second example we declare that whatever is returned from fold [], it should be a String. From that the compiler infers that [] actually must have a type of [String]. Last fold is the simplest: the program folds over elements in list and concatenates them because concatenation is the operation defined in Monoid String typeclass instance.

Back to options (or more precisely Maybe). Folding over Maybe monad having type parameter being Monoid (I can't believe I just said it) has an interesting interpretation: it either returns value inside Maybe or a default Monoid value:

> fold (Just "abc")
"abc"
> fold Nothing :: String
""

Just "abc" is same as Some("abc") in Scala. You can see here that if Maybe String is Nothing, neutral String monoid value is returned, that is an empty string.

Summary

Haskell shows that folding (also over Maybe) can be at least consistent. In Scala Option.fold is unrelated to List.fold, confusing and unreadable. I advise avoiding it and staying with slightly more verbose map/getOrElse transformations or pattern matching.

Comments

Scala Puzzlers #25 shows the problem with getOrElse, a run-time error. With fold, you get a compile time error. Scala's strong type safety is a strength. Granted a name like fold2 or something else might have been better. The drawback of looking at negative case first is really really so minor when compared to type checking advantage. Code from Puzzlers below: